Given, $x=2 t /\left(1+t^{2}\right)$
On differentiating $x$ with respect to t using quotient rule,
$$
\begin{array}{l}
\frac{\mathrm{dx}}{\mathrm{dt}}=\left[\frac{\left(1+\mathrm{t}^{2}\right) \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{t})-2 \mathrm{t} \frac{\mathrm{d}}{\mathrm{dt}}\left(1+\mathrm{t}^{2}\right)}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \\
=\left[\frac{\left(1+\mathrm{t}^{2}\right)(2)-2 \mathrm{t}(2 \mathrm{t})}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \\
=\left[\frac{2+2 \mathrm{t}^{2}-4 \mathrm{t}^{2}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \\
=\left[\frac{2-2 \mathrm{t}^{2}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \\
\frac{\mathrm{dx}}{\mathrm{dt}}=\left[\frac{2-2 \mathrm{t}^{2}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \\
\mathrm{y}=\left(1-\mathrm{t}^{2}\right) /\left(1+\mathrm{t}^{2}\right)
\end{array}
$$
On differentiating y with respect to t using quotient rule,
$\frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}\left(1-t^{2}\right)-\left(1-t^{2}\right) \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$
$\begin{array}{l}
=\left[\frac{\left(1+\mathrm{t}^{2}\right)(-2 \mathrm{t})-\left(1-\mathrm{t}^{2}\right)(2 \mathrm{t})}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \\
=\left[\frac{-2 \mathrm{t}-2 \mathrm{t}^{3}-2 \mathrm{t}+2 \mathrm{t}^{3}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \\
\frac{\mathrm{dy}}{\mathrm{dt}}=\left[\frac{-4 \mathrm{t}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \ldots \ldots .(2)
\end{array}$
On dividing equation (2) by (1)
$\begin{array}{l}
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\left[\frac{-4 \mathrm{t}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \times \frac{1}{\left[\frac{2-2 \mathrm{t}^{2}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right]} \\
=-\frac{2 \mathrm{t}}{1-\mathrm{t}^{2}} \\
\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\mathrm{y}}\left[\sin \mathrm{ce}, \frac{\mathrm{x}}{\mathrm{y}}=\frac{2 \mathrm{t}}{1+\mathrm{t}^{2}} \times \frac{1+\mathrm{t}^{2}}{1-\mathrm{t}^{2}}=\frac{2 \mathrm{t}}{1-\mathrm{t}^{2}}\right] \\
\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\mathrm{y}}\left[\sin \mathrm{se}, \frac{\mathrm{x}}{\mathrm{y}}=\frac{2 \mathrm{t}}{1+\mathrm{t}^{2}} \times \frac{1+\mathrm{t}^{2}}{1-\mathrm{t}^{2}}=\frac{2 \mathrm{t}}{1-\mathrm{t}^{2}}\right]
\end{array}$